Question

find the average marks of the students from the following table using assumed mean method.2nd question

Answer 1

Answer:

Average marks= 50

Step-by-step explanation:

First convert the given more than CF table to simple class interval table

class interval--freq--class mark--xifi

0_10_______5________5_______25

10-20______5________15______75

20-30______5_______25_____125

30-40______12______35_____420

40-50______9______45______405

50-60_____21______55_____1155

60-70_____7_______65______455

70-80_____7_______75_______525

80-90_____4_______85_______340

90-100_____5______95_______475

Direct mean

$\bar{x} = \frac{ \Sigma \: x_if_i}{\Sigma \: f_i} \\ \\ \bar{x} = \frac{25 + 75 + 125 + 420 +405 + 1155 + 455 + 525 +340 + 475 }{80} \\ \\ = \frac{4000}{80} \\ \\ \bar{x} = 50$

Average marks of students = 50

Assume Mean:

a= 45

CI_______fi_____xi____di_____difi

0-10____5_____5_____-40____-200

10-20___5_____15___ -30_____-150

20-30___5_____25___ -20____-100

30-40___12____35___-10____-120

40-50___9_____45____0______0

50-60___21____55___10_____210

60-70___7_____65___20_____140

70-80___7_____75____30____210

80-90___4_____85____40____160

90-100___5____95____50____250

$\bar{x} = a + \frac{ \Sigma \: d_if_i}{\Sigma \: f_i} \\ \\ \bar{x} = 45 + \frac{ - 200 - 150 - 100 - 120 + 0 +210 + 140 + 210 +160 + 250 }{80} \\ \\ = 45 + \frac{ - 570 + 970}{80} \\ \\ \bar{x} = 45 + \frac{400}{80} \\ \\ \bar{x} = 45 + 5 \\ \\ \bar{x} = 50$

Hope it helps you.

Answer 2

Answer:

Average marks of students = 50

Assume Mean:

a= 45

CI_______fi_____xi____di_____difi

0-10____5_____5_____-40____-200

10-20___5_____15___ -30_____-150

20-30___5_____25___ -20____-100

30-40___12____35___-10____-120

40-50___9_____45____0______0

50-60___21____55___10_____210

60-70___7_____65___20_____140

70-80___7_____75____30____210

80-90___4_____85____40____160

90-100___5____95____50____250

\begin{gathered}\bar{x} = a + \frac{ \Sigma \: d_if_i}{\Sigma \: f_i} \\ \\ \bar{x} = 45 + \frac{ - 200 - 150 - 100 - 120 + 0 +210 + 140 + 210 +160 + 250 }{80} \\ \\ = 45 + \frac{ - 570 + 970}{80} \\ \\ \bar{x} = 45 + \frac{400}{80} \\ \\ \bar{x} = 45 + 5 \\ \\ \bar{x} = 50 \\ \\ \end{gathered}

x

ˉ

=a+

Σf

i

Σd

i

f

i

x

ˉ

=45+

80

−200−150−100−120+0+210+140+210+160+250

=45+

80

−570+970

x

ˉ

=45+

80

400

x

ˉ

=45+5

x

ˉ

=50